A machine with efficiency 2 / 3 used 12 J of energy in lifting 2 kg block through certain height and it is allowed to fall through the same. The velocity while it reach the ground is

Step-by-step solution

Potential energy of the machine, aligned & U = 2 3 12=8 ~J & mgh =8 ~h = 8 2 10 =0.4 ~m aligned By conservation of energy, Loss in PE = Gain in K.E aligned & mgh = 1 2 mv ^2 & v = 2 gh = 2 10 0.4 =2 2 ~m / s aligned